**Reaction at ***B*

$\Sigma M_E = 0$

$7.5R_B = 3(1.5) + 4.5(3) + 2.25(3)$

$R_B = 3.3 ~ \text{kN}$

**Section Through ***M-M*

$\Sigma F_V = 0$

$CF_V = R_B$

$CF \sin 45^\circ = 3.3$

$CF = 3.3\sqrt{2} ~ \text{kN tension}$

$CF = 4.67 ~ \text{kN tension}$ ← [ C ] *answer for part 1*

**At Joint ***B*

$\Sigma F_V = 0$

$BF = R_B$

$BF = 3.3 ~ \text{kN compression}$ ← [ B ] *answer for part 2*

**Section Through ***N-N*

$\Sigma F_V = 0$

$DG_V + 3 = R_B$

$DG(4/5) + 3 = 3.3$

$DG = 0.375 ~ \text{kN tension}$

**From Joint ***D*

$\Sigma F_V = 0$

$DH + (4/5)DG = 3$

$DH + (4/5)(0.375) = 3$

$DH = 2.7 ~ \text{kN tension}$ ← [ D ] *answer for part 3*